Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is: [7]
[2, 2, 3]
DFS!
1 class Solution { 2 public: 3 void findNext(vector<vector<int> > &res, vector<int> &candidates, vector<int> v, int sum, int target, int idx) { 4 if (sum > target || idx >= candidates.size()) { 5 return; 6 } 7 if (sum == target) { 8 res.push_back(v); 9 return; 10 } 11 v.push_back(candidates[idx]); 12 findNext(res, candidates, v, sum + candidates[idx], target, idx); 13 v.pop_back(); 14 findNext(res, candidates, v, sum, target, idx + 1); 15 } 16 17 vector<vector<int> > combinationSum(vector<int> &candidates, int target) { 18 vector<vector<int> > res; 19 sort(candidates.begin(), candidates.end()); 20 vector<int> v; 21 findNext(res, candidates, v, 0, target, 0); 22 return res; 23 } 24 };